What is the common preimage (in $Z$) and the equivalence relation for Pushouts Here it says: Suppose that $X$, $Y$, and $Z$ as above are sets, and that $f : Z → X$ and $g : Z → Y$ are set functions. The pushout of $f$ and $g$ is the disjoint union of $X$ and $Y$, where elements sharing a common preimage (in $Z$) are identified, together with certain morphisms from X and Y.
I do not understand this part "the pushout of $f$ and $g$ is the disjoint union of $X$ and $Y$, where elements sharing a common preimage (in $Z$) are identified". Question: To form the Pushout you get the disjoint union and then what?
 A: First I suppose we should formalise disjoint union; I'll say
$$X \sqcup Y = (X \times \{ 0 \}) \cup (Y \times \{ 1 \})$$
so that each $x \in X$ corresponds with $(x,0) \in X \sqcup Y$ and each $y \in Y$ corresponds with $(y,1) \in X \sqcup Y$. The $0$ and $1$ are redundant apart from the fact that they ensure that the components of the above union are disjoint even if $X$ and $Y$ are not disjoint... imagine laying the set $X$ out on the ground (level $0$) and laying the set $Y$ out upstairs (level $1$).
We can define the pushout of $f$ and $g$ in the category of sets to be the set
$$P = (X \sqcup Y) / {\sim}$$
where $\sim$ is the smallest equivalence relation satisfying $(f(z),0) \sim (g(z),1)$ for all $z \in Z$. That is, $\sim$ takes the disjoint union and glues together points downstairs in $X$ and points upstairs in $Y$: it identifies two points if and only if there is some point in $Z$ that they both come from.
To put it another way, $\sim$ is the least equivalence relation such that $(x,0) \sim (y,1)$ whenever $f^{-1} (\{ x \}) \cap g^{-1} (\{ y \}) \ne \varnothing$.

Example. Let $Z = \{ a,b \}$ and let $X = Y = \mathbb{Z}$. Define $f : \{ a,b \} \to \mathbb{Z}$ and $g : \{ a,b \} \to \mathbb{Z}$ by
$$f(a) = 5, \quad f(b) = 6,$$
$$g(a) = 7, \quad g(b) = 8$$
Now the disjoint union of $X$ and $Y$ is just two copies of $\mathbb{Z}$, i.e.
$$X \sqcup Y = \{ (n,i) : n \in \mathbb{Z},\ i \in \{ 0,1 \} \}$$
The equivalence relation is going to identify every element with itself, together with
$$(5,0) \sim (7,1) \quad \text{and} \quad (6,0) \sim (8,1)$$
since $5 = f(a)$ and $7=g(a)$, and $6=f(b)$ and $8=g(b)$.
The pushout is then $\mathbb{Z} \cup \mathbb{Z}$, except with $(5,0)$ and $(7,1)$ regarded as the same element, and $(6,0)$ and $(8,1)$ regarded as the same element (since they lie in the same $\sim$-equivalence class).

Another example. Let $X=Y=S^1 \subseteq \mathbb{R}^2$ be unit circles in the plane and let $Z = \{ a \}$. Define $f(a)=g(a)=(1,0)$. Thus $f$ and $g$ just pick out a point on the circle.
From a geometric perspective, $X \sqcup Y$ is what you get when you lie two circles next to each other in such a way that they don't overlap. The pushout of $f$ and $g$ identifies $f(a) \in X$ with $g(a) \in Y$, i.e. it joins together the two points.
The pushout is thus a figure-eight shape.
